Puzzle for February 27, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
* CD is a 2-digit number (not C×D).
Scratchpad
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Hint #1
Since D ≠ 0 (from eq.5), divide both sides of eq.1 by D: (B × C) ÷ D = D ÷ D which becomes (B × C) ÷ D = 1 Divide both sides of the above equation by B (assumes B ≠ 0): ((B × C) ÷ D) ÷ B = 1 ÷ B which becomes eq.1a) C ÷ D = 1 ÷ B
Hint #2
Confirm: B ≠ 0 ... If B = 0, then replacing B with 0 in eq.1 would yield: eq.1) 0 × C = D which would make 0 = D Since D ≠ 0 (from eq.5), then B ≠ 0
Hint #3
In eq.1a, replace C ÷ D with A – F (from eq.5): eq.1b) A – F = 1 ÷ B To make eq.1b true, check several possible values for B: If B = 1, then A – F = 1 ÷ 1 = 1 If B = 2, then A – F = 1 ÷ 2 = ½ If B > 2, then 0 < A – F < ½ Since A and F are integers, then A – F must be an integer, which means: eq.1c) A – F = 1 which makes B = 1
Hint #4
In eq.1, substitute 1 for B: 1 × C = D which makes C = D
Hint #5
Add F to both sides of eq.1c: A – F + F = 1 + F which makes eq.1d) A = 1 + F Substitute (1 + F) for A (from eq.1d), and 1 for B in eq.4: (1 + F) × 1 = C + F which becomes 1 + F = C + F Subtract F from both sides of the equation above: 1 + F – F = C + F – F which makes 1 = C and also makes D = C = 1
Hint #6
eq.3 may be written as: E × F = A + 10×C + D In the equation above, substitute 1 + F for A (from eq.1d), and 1 for C and D: E × F = 1 + F + 10×1 + 1 which becomes E × F = F + 12 Divide both sides by F: E × F ÷ F = (F + 12) ÷ F which makes eq.3a) E = (F + 12) ÷ F
Hint #7
Substitute (1 + F) for A (from eq.1d), ((F + 12) ÷ F) for E (from eq.3a), and 1 for B in eq.2: (1 + F) × F = ((F + 12) ÷ F) – 1 Add 1 to both sides of the equation above: (1 + F) × F + 1 = ((F + 12) ÷ F) – 1 + 1 which becomes F + F² + 1 = (F + 12) ÷ F Multiply both sides by F: (F + F² + 1) × F = (F + 12) ÷ F × F which becomes F² + F³ + F = F + 12 Subtract F from both sides: F² + F³ + F – F = F + 12 – F which becomes eq.2a) F² + F³ = 12
Solution
To make eq.2a true, check several possible values for F: If F = 0, then F² + F³ = 0² + 0³ = 0 + 0 = 0 If F = 1, then F² + F³ = 1² + 1³ = 1 + 1 = 2 If F = 2, then F² + F³ = 2² + 2³ = 4 + 8 = 12 If F = 3, then F² + F³ = 3² + 3³ = 9 + 27 = 36 If F > 3, then F² + F³ > 36 Since F² + F³ = 12, the above equations make: F = 2 making A = 1 + F = 1 + 2 = 3 (from eq.1d) E = (F + 12) ÷ F = (2 + 12) ÷ 2 = 14 ÷ 2 = 7 (from eq.3a) and ABCDEF = 311172